从成对增加的相邻数组元素中重复移除一个元素后，找到最后剩余的元素
原文:https://www . geeksforgeeks . org/find-从递增相邻数组元素对中重复移除元素后最后剩余的元素/
给定一个由 N 个整数组成的数组 arr[] ，任务是在重复选择递增的相邻元素对 (arr[i]，arr[i + 1]) 并移除对中的任何元素后，打印最后剩余的数组元素。如果无法将数组缩减为单个元素，则打印“不可能”。
示例:
输入: arr[] = {3，1，2，4} 输出: 4 解释: 第一步:选择一对(1，2)，从中去掉 2。现在数组变成了[3，1，4]。 第二步:选择配对(1，4)并从中移除 1。现在数组变成了[3，4]。 第三步:选择配对(3，4)并从中移除 3。现在数组变成了[4]。 因此，剩余元素为 4。
输入: arr[] = {2，3，1} 输出:不可能
方法:给定的问题可以通过观察这样一个事实来解决:如果第一个数组元素小于最后一个数组元素，那么它们之间的所有元素都可以通过执行给定的操作来移除。因此，只需检查是否 arr[0]<arr[N–1]。如果发现为真，打印第一个或最后一个数组元素。否则，打印 -1 。
下面是上述方法的实现:
C++14
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the last remaining
// array element after after performing
// given operations
void canArrayBeReduced(int arr[], int N)
{
// If size of the array is 1
if (N == 1) {
cout << arr[0];
return;
}
// Check for the condition
if (arr[0] < arr[N - 1]) {
cout << arr[N - 1];
}
// If condition is not satisfied
else
cout << "Not Possible";
}
// Driver Code
int main()
{
int arr[] = { 6, 5, 2, 4, 1, 3, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
canArrayBeReduced(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
class GFG
{
// Function to print the last remaining
// array element after after performing
// given operations
static void canArrayBeReduced(int[] arr, int N)
{
// If size of the array is 1
if (N == 1)
{
System.out.print(arr[0]);
return;
}
// Check for the condition
if (arr[0] < arr[N - 1])
{
System.out.print(arr[N - 1]);
}
// If condition is not satisfied
else
System.out.print("Not Possible");
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 6, 5, 2, 4, 1, 3, 7 };
int N = arr.length;
// Function Call
canArrayBeReduced(arr, N);
}
}
// This code is contributed by Dharanendra L V.
Python 3
# Python 3 program for the above approach
# Function to print the last remaining
# array element after after performing
# given operations
def canArrayBeReduced(arr, N):
# If size of the array is 1
if (N == 1):
print(arr[0])
return
# Check for the condition
if (arr[0] < arr[N - 1]):
print(arr[N - 1])
# If condition is not satisfied
else:
print("Not Possible")
# Driver Code
if __name__ == "__main__":
arr = [6, 5, 2, 4, 1, 3, 7]
N = len(arr)
# Function Call
canArrayBeReduced(arr, N)
# This code is contributed by chitranayal.
C
// C# program for the above approach
using System;
public class GFG {
// Function to print the last remaining
// array element after after performing
// given operations
static void canArrayBeReduced(int[] arr, int N)
{
// If size of the array is 1
if (N == 1)
{
Console.Write(arr[0]);
return;
}
// Check for the condition
if (arr[0] < arr[N - 1])
{
Console.Write(arr[N - 1]);
}
// If condition is not satisfied
else
Console.Write("Not Possible");
}
// Driver Code
static public void Main()
{
int[] arr = { 6, 5, 2, 4, 1, 3, 7 };
int N = arr.Length;
// Function Call
canArrayBeReduced(arr, N);
}
}
// This code is contributed by Dharanendra L V.
java 描述语言
<script>
// JavaScript program for the above approach
// Function to print the last remaining
// array element after after performing
// given operations
function canArrayBeReduced(arr, N)
{
// If size of the array is 1
if (N == 1)
{
document.write(arr[0]);
return;
}
// Check for the condition
if (arr[0] < arr[N - 1])
{
document.write(arr[N - 1]);
}
// If condition is not satisfied
else
document.write("Not Possible");
}
// Driver Code
let arr = [ 6, 5, 2, 4, 1, 3, 7 ];
let N = arr.length;
// Function Call
canArrayBeReduced(arr, N);
// This code is contributed by Surbhi Tyagi.
</script>
Output:
7
时间复杂度:O(1) T5辅助空间:** O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
